Posted by Jordan on Saturday, September 12, 2009 at 1:21pm.
A shotputter throws the shot (mass = 7.3kg) with an initial speed of 15.0 m/s at a 33.0 degree angle to the horizontal.
Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.00m above the ground.

Physics  bobpursley, Saturday, September 12, 2009 at 3:30pm
no air resistance?
consider the horizonal distance:
d= Vihorizontal*time = 15cos33*t
now the vertical:
finalheight=initialheight+ vivertical*t 4.9t^2
or
0=2+15sin33 t  4.9 t^2
solve the second equation for time. Use the quadratic equation. Put that time t into the first equation (distance horizontal). 
Physics  Anonymous, Tuesday, January 15, 2013 at 9:47pm
sdds